Precalculus (6th Edition) Blitzer

a) The value is, $\frac{\sqrt{2}}{2}$. b) The value is, $\frac{\sqrt{2}}{2}$.
(a) In the given unit circle, the point that corresponds to $t=\frac{3\pi }{4}$ has the coordinates $\left( -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ So use $x=-\frac{\sqrt{2}}{2}$ and $y=\frac{\sqrt{2}}{2}$. Such that, $\sin \left( \frac{3\pi }{4} \right)=y=\frac{\sqrt{2}}{2}$ The value of the trigonometric function $\sin \left( \frac{3\pi }{4} \right)$ is $\frac{\sqrt{2}}{2}$. (b) We have the periodic properties of sine and cosine functions: $\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$. Therefore, \begin{align} & \sin \left( \frac{11\pi }{4} \right)=\sin \left( \frac{3\pi }{4}+2\pi \right) \\ & =\sin \left( \frac{3\pi }{4} \right) \end{align} Now, $\sin \left( \frac{3\pi }{4} \right)=\frac{\sqrt{2}}{2}$. So, $\sin \left( \frac{3\pi }{4} \right)=\frac{\sqrt{2}}{2}.$ Thus, the value of the trigonometric function $\sin \left( \frac{11\pi }{4} \right)$ is, $\frac{\sqrt{2}}{2}$.